Evaluate $~~\int\dfrac{\ln x}{x^2}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\ln x}x+\dfrac1{x^2}+C$ (Choice B) B $-\dfrac{\ln x}x-\dfrac2{x^2}+C$ (Choice C) C $-\dfrac{\ln x}x+\dfrac1x+C$ (Choice D) D $-\dfrac{\ln x}x-\dfrac1x+C$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv= \dfrac1{x^2}\,dx\,$. Then $~du = \dfrac1xdx~$ and $~v = -\dfrac1x\,$. Integration by parts gives $ \int \dfrac{\ln x}{x^2}dx =-\dfrac{\ln x}x+\int\dfrac1{x^2}dx$ $ \,=-\dfrac{\ln x}x-\dfrac1x+C\,$.